From: Mark Wilson <mwilson13@...>
Date: 2003-10-01T11:16:20+09:00
Subject: Re: Making == symmetric?

Those more knowledgeable than me should correct what I've written below 
if it's wrong.

On Tuesday, September 30, 2003, at 07:19 PM, Nathan Weston wrote:

> It has always bothered me that == is not symmetric in ruby:
> a == b is shorthand for a.==(b), while b == a is shorthand for
> b.==(a), and a and b might not agree on whether they are equal.

I think it is symmetric so long as you bear in mind that the method 
belongs to the object and not to the variable that refers to the 
object. See below.

> [snip]

> First of all: yes, this issue does come up in real code.
> Here's a quick example I ran into today:
>
> require "delegate"
> class Foo
> end
> class D < SimpleDelegator
>   def initialize(obj)
>     super(obj)
>   end
> end
> f = Foo.new
> d = D.new(f)
> d == f  #evaluates to true
> f == d  #evaluates to false

I think you're mixing up variables and values. d refers to a D object 
instantiated with a Foo object. f refers to the Foo object used to 
instantiate the D object. The D object has an == method that checks its 
value and compares it to the given value. When invoked by d==f, the D 
object sees that its value is the Foo object and sees that f refers to 
the Foo object and determines that they are the same object. In other 
words the D object is the functional equivalent of an assignment to a 
primitive variable.

The Foo object does not have a value other than its object id (I 
think). When it queries the value of the variable d in performing the 
== method, it gets the object id of the D object (I think). The two are 
not equal. The proper comparison would be:

f.inspect==d.inspect

Another way of looking at it:

D.new(Foo.new) == Foo.new # false
Foo.new == D.new(Foo.new) # false



> I ran into a similar problem trying to implement perl6-style
> junctions: you end up with any(1,2,3) == 1 being true, but 1 ==
> any(1,2,3) being false.

I didn't follow the above point. There was an earlier discussion on the 
list about perl6-style junctions that might be helpful.

> [snip]

Regards,

Mark